The majority of text books that I looked at cover the topic of exponential
functions from a purely algebraic approach. This makes sense since this
topic is introduced in an high school algebra II class. Applications are
covered in the traditional text, some more than others. What I have attempted
to do is create a plan that uses applications to introduce these valuable
functions.

GROWTH APPLICATIONS

Mathematics can be used to model certain phenomenon that exist in the
real world. We create these models to help us predict the future. One of
the functions the mathematicians use is called an exponential function.
Exponential functions are of the form: f(x) = A^x , where A is the base
and X is the exponent.

The first problem we should look at deals with population
growth.

If we look at what makes a population grow over one year maybe we can develop
a model that will help us predict the future.

Suppose that the population of a certain town is P is increasing at a constant
rate of R each year. We can see that the population of our town one year
later will be:

Now we have a model that we can use to predict the future
population of a particular town so the local government will know how to
plan for the future growth of the community.

For example, a town has a population of 50,000 that is increasing at a rate
of 2.5% each year. Use the model above to predict the population in 10 years.

The first thing we need to do is fill in the correct numbers from our problem
into our model.

P = 50,000 people
R = 2.5% or .025
t = 10 years

Our model looks like this:

50,000 * ( 1 + .025 )^ 10 = Y ; add the numbers within ( )

50,000 * (1.025)^ 10 = Y ; use calc. to evaluate (1.025)^ 10

50,000 * ( 1.28 ) = 64,004 ; multiply

So the population of our town will be 64,004 in ten years. What would
you do if we needed to know when the population of our town will double?
Why don't we use the same approach we used for the first problem. Let's
fill in our model!!!

At this time we don't have the algebra skills to solve this exponential
function. Could we use our TI-82 to solve for t ?

The answer is YES !!!!!

Enter the equation (1.025)^ t = 2 into the Y= function of your calculator.
Now go to the Table function of your calculator and scroll down until the
Y1 column reads around 100,000. You can make your answer more accurate by
adjusting TblSet on your calculator.

Now use your TRACE function to scroll to where Y is approximately 100,000
and see if X is about 28.

The graph should look like this algebra Expressor file.

Several types of growth can be modeled
using Exponential functions. For example, assume that a certain type of
bacterial doubles every hour and there are initially 100 organisms. Lets
develop a model for the bacterial growth.

initial amount /////// \\\\\\\one hour later /////// \\\\\\\ two hours later
/////// \\\\\\\ three hours later

So the model we have developed is Y = 100 * ( 2 )^ X , where Y is the new
amount of bacterial and X is the number of hours later.Let's use this model to solve the following
problem. Suppose that the above bacterial will be used in an experiment
that allows the bacterial to grow for one day. How many organisms will there
be at the end of the experiment if the scientists stared with 100 organisms?

Now we need to change the problem so we are
looking for the amount of time it will take our bacteria to reach a specific
amount of organisms.

What if scientist discovered that our bacteria could be used to create a
cure for the common cold. The only catch to this remarkable development
is that 200,000 organisms are needed to create one dose of the new miracle
drug. If the scientist started with 25 organisms, how long would it take
for the bacteria to grow enough to make 100 doses of the new medicine?

If we look at the previous problem we can see that we started with 100 organisms.
This problem starts with 25 organisms, so our model becomes Y = 25 * ( 2
)^ X . Now all we have to do is fill in our model.

Everyone is interested in money. So let's take some time to look at the
some financial applications of exponential functions. Assume that P dollars
are invested at a simple interest rate of R, then P is referred to as the
Principle and P * R is the amount of interest earned at the end of one period.
Now we should be able to develop a model.

So now that we have a model, let's use it to find out how much money
we will have in the future!!!

Max deposited $750 in an account that pays
7% simple annual interest. How much money will Max have at the end of five
years?

Given our model P ( 1 + N * R ) = T , P = $750 R = .07 N = 5

Simply plug and chug to arrive at the answer.

750 * ( 1 + ( 5 * .07 )) = T ; evaluate on your calculator

1012.5 = T

Wait where is the exponent ? There isn't one,
so this isn't an exponential function. Why did we look at this function?
Well , most banks don't offer simple interest. What they do is allow interest
to compound ; this is to your advantage. When a bank compounds interest
they are paying interest on interest.

So we need to create a model for compounding interest. Suppose that we invest
P dollars at a rate R compounded 4 times per year, that is called compounding
quarterly. We should expect to receive 1/4 of the interest received for
one year each quarter. So the simple interest rate in this case where
R = 7% is .07/4.

On the flip side of this we need to see that N becomes the number of times
we calculated interest earned. So if we had a problem where interest is
compounded quarterly over a 10 year period

N = 4 * 10 = 40

Now our model will take the form:

T = P * ( 1 + ( R/k))^ N

where k = the number of times interest is compounded each year and N
= the total number of times interest is compounded.

Let's look at an example.

Max deposited $750 in an account that pays 7% interest compounded quarterly.
How much money will Max have at the end of five years?

If Drew invested $1550 in a bank that offers a rate of 8.5% compounded quarterly
for 7 years and Craig invested $1550 in a different bank at a rate of 8%
compounded monthly for 7 years. Who would have more money at the end of
7 years?

This is two problems that we can evaluate using our model and compare.

Consider the following problem. Craig needs
$5000 to buy his uncle's 1967 Corvette Stingray and his uncle has agreed
to keep the car until Craig has the money. If Craig has $1200 to invest
at a rate of 9% compounded monthly, how long will it take for Craig to get
his dream car?

We should enter the original equation , Y1 = 1200 * ( 1 + (.09/12))Y
* 12 , into your calculator and go to the TABLE function or the graph and
find the approximation for Y.

CLICK HERE to see the Excel spreadsheet that
I used to solve this problem.

This is an Algebra Expressor file I used to check my answer.

Another application to investigate how
the total value of an investment changes as the number of times interest
is compounded increases. If $1000 is invested at 7% for one year, what is
the difference between the total amounts of the investment if the number
of times interest is compounded increases.

Compounding ..........................Total value after one year
period

As we can see, this seems to approach some limit as the number of times
we compound interest approaches infinity.

When we let the number of times interest is compounded be infinity we say
that interest is being compounded continuously. The model that results from
this limit is Total = Pe^ RT , where P = the principle , R = the given rate
of interest, T = the number of years.

The value of e = 2.718281828........ can be found by evaluating the following
summation ;

e = 1 + (1/1!) + (1/2!) + (1/3!) + (1/4!) + ............... +
(1/n!).

This is a bit complicated , but can be evaluated by most students with
a little guidance.

If you would like to see the Excel spread sheet I used to calculate e CLICK HERE!!!

Lets use our new model for computing the value
of an investment when interest is compounded continuously. If Max invests
$500 at 7.5% interest compounded continuously, how long will it take for
his money to double?

Enter Y1 = e ^ (.075 T) into the Y= function and use either your table
function or the graph to find an approximation. The answer is somewhere
between 9 and 10 years.

If you would like to see the Excel spread sheet In used to solve this problem
CLICK HERE!!!

I have successfully avoided the Algebra that
is used to introduce exponential functions. The algebra is important but
the applications of mathematics is the part that interests most people and
I think most students in high school. Obviously the properties of exponential
functions need to be seen. As well as logarithmic functions and their properties,
so students can gain the ability to solve problems when the technology is
not accessible. Wait I can carry my TI-82 everywhere, do think the students
have figured this out already?